package Leetcode100Hot;

import java.util.ArrayList;
import java.util.List;

/*
二叉树展开为链表
给你二叉树的根结点 root ，请你将它展开为一个单链表：
展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
展开后的单链表应该与二叉树 先序遍历 顺序相同。

示例 1：
输入：root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]
示例 2：
输入：root = []
输出：[]
示例 3：
输入：root = [0]
输出：[0]

提示：
树中结点数在范围 [0, 2000] 内
-100 <= Node.val <= 100

进阶：你可以使用原地算法（O(1) 额外空间）展开这棵树吗？
 */
public class _75二叉树展开为链表 {

    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode() {
        }

        TreeNode(int val) {
            this.val = val;
        }

        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }

    //方法一：前序遍历
    /*
    作者：力扣官方题解
    链接：https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/solutions/356853/er-cha-shu-zhan-kai-wei-lian-biao-by-leetcode-solu/
     */
    class Solution {
        public void flatten(TreeNode root) {
            List<TreeNode> list = new ArrayList<TreeNode>();
            preorderTraversal(root, list);
            int size = list.size();
            for (int i = 1; i < size; i++) {
                TreeNode prev = list.get(i - 1), curr = list.get(i);
                prev.left = null;
                prev.right = curr;
            }
        }

        public void preorderTraversal(TreeNode root, List<TreeNode> list) {
            if (root != null) {
                list.add(root);
                preorderTraversal(root.left, list);
                preorderTraversal(root.right, list);
            }
        }
    }

    //官解：方法三：寻找前驱节点
    //思路无敌，O(n) O(1)
    /*
    作者：力扣官方题解
    链接：https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/solutions/356853/er-cha-shu-zhan-kai-wei-lian-biao-by-leetcode-solu/
     */
    class Solution2 {
        public void flatten(TreeNode root) {
            TreeNode curr = root;
            while (curr != null) {
                if (curr.left != null) {
                    TreeNode next = curr.left;
                    TreeNode predecessor = next;
                    while (predecessor.right != null) {
                        predecessor = predecessor.right;
                    }
                    predecessor.right = curr.right;
                    curr.left = null;
                    curr.right = next;
                }
                curr = curr.right;
            }
        }
    }

}
